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I have worked this problem at least 30 times and am still not getting the correct answer. Can anyone tell me where I'm wrong?

Partial sum of a geometric series:

enter image description here $$\sum_{n=1}^{\infty} \frac{8^n+3^n}{9^n}$$

for the sum formula $\frac{a}{1-r}$ I have the values for the variables as:

$a = \frac{11}{9}$

$r = \frac{73}{99}$

I found $r$ by dividing the first value of the series ($\frac{11}{9}$) by the second value of the series ($\frac{73}{81}$).

I keep getting the sum as $\frac{121}{26}$

The correct value for the sum is $\frac{17}{2}$

What am I doing wrong?

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  • $\begingroup$ The series definition is missing.... Says "enter image description here" $\endgroup$
    – Brick
    Commented Jun 7, 2019 at 17:07

3 Answers 3

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You will have $$\sum_{n=1}^\infty\left(\frac{8}{9}\right)^n$$ and $$\sum_{n=1}^\infty\left(\frac{1}{3}\right)^n$$ The first sum is $$8$$ and the second $$\frac{1}{2}$$

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  • $\begingroup$ Yes, thank you so much! $\endgroup$
    – user680414
    Commented Jun 8, 2019 at 20:44
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I don't really understand what you did, but the formula for the sum of the geometric series starting from exponent $1$ is $$\sum_{n=1}^\infty x^n=x+x^2+\dots=\frac x{1-x}\qquad(|x|<1)$$

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  • $\begingroup$ He blindly applied the formula as if it was a single geometric series. $\endgroup$
    – user65203
    Commented Jun 8, 2019 at 10:39
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You must handle the two series separately.

$$\sum_{n=1}^{\infty} \frac{8^n}{9^n}=\frac8{9\left(1-\dfrac89\right)}=8,$$

$$\sum_{n=1}^{\infty} \frac{3^n}{9^n}=\frac1{3\left(1-\dfrac13\right)}=\frac12.$$

Then add both.

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  • $\begingroup$ Thank you! That was it, I feel like such a dummy but now I can move forward :) $\endgroup$
    – user680414
    Commented Jun 8, 2019 at 20:43

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